Sum of previous numbers that are greater than current number for given array
Given an array A[], for each element in the array, the task is to find the sum of all the previous elements which are strictly greater than the current element.
Examples:
Input: A[] = {2, 6, 4, 1, 7}
Output: 0 0 6 12 0
Explanation:
For 2 and 6 there is no element greater to it on the left.
For 4 there is 6.
For 1 the sum would be 12.
For 7 there is again no element greater to it.Input: A[] = {7, 3, 6, 2, 1}
Output: 0 7 7 16 18
Explanation:
For 7 there is no element greater to it on the left.
For 3 there is 7.
For 6 the sum would be 7.
For 2 it has to be 7 + 3 + 6 = 16.
For 1 the sum would be 7 + 3 + 6 + 2 = 18
Naive Approach: For each element, the idea is to find the elements which are strictly greater than the current element on the left side of it and then find the sum of all those elements.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Max Element of the Array const int maxn = 1000000; // Function to find the sum of previous // numbers that are greater than the // current number for the given array void sumGreater( int ar[], int N) { // Loop to iterate over all // the elements of the array for ( int i = 0; i < N; i++) { // Store the answer for // the current element int cur_sum = 0; // Iterate from (current index - 1) // to 0 and check if ar[j] is greater // than the current element and add // it to the cur_sum if so for ( int j = i - 1; j >= 0; j--) { if (ar[j] > ar[i]) cur_sum += ar[j]; } // Print the answer for // current element cout << cur_sum << " " ; } } // Driver Code int main() { // Given array arr[] int ar[] = { 7, 3, 6, 2, 1 }; // Size of the array int N = sizeof ar / sizeof ar[0]; // Function call sumGreater(ar, N); return 0; } |
Java
// Java program for the above approach class GFG{ // Max Element of the Array static int maxn = 1000000 ; // Function to find the sum of previous // numbers that are greater than the // current number for the given array static void sumGreater( int ar[], int N) { // Loop to iterate over all // the elements of the array for ( int i = 0 ; i < N; i++) { // Store the answer for // the current element int cur_sum = 0 ; // Iterate from (current index - 1) // to 0 and check if ar[j] is greater // than the current element and add // it to the cur_sum if so for ( int j = i - 1 ; j >= 0 ; j--) { if (ar[j] > ar[i]) cur_sum += ar[j]; } // Print the answer for // current element System.out.print(cur_sum + " " ); } } // Driver Code public static void main(String[] args) { // Given array arr[] int ar[] = { 7 , 3 , 6 , 2 , 1 }; // Size of the array int N = ar.length; // Function call sumGreater(ar, N); } } // This code is contributed by amal kumar choubey |
Python3
# Python3 program for the above approach # Max Element of the Array maxn = 1000000 ; # Function to find the sum of previous # numbers that are greater than the # current number for the given array def sumGreater(ar, N): # Loop to iterate over all # the elements of the array for i in range (N): # Store the answer for # the current element cur_sum = 0 ; # Iterate from (current index - 1) # to 0 and check if ar[j] is greater # than the current element and add # it to the cur_sum if so for j in range (i, - 1 , - 1 ): if (ar[j] > ar[i]): cur_sum + = ar[j]; # Print the answer for # current element print (cur_sum, end = " " ); # Driver Code if __name__ = = '__main__' : # Given array arr ar = [ 7 , 3 , 6 , 2 , 1 ] ; # Size of the array N = len (ar); # Function call sumGreater(ar, N); # This code is contributed by sapnasingh4991 |
C#
// C# program for the above approach using System; class GFG{ // Max Element of the Array //static int maxn = 1000000; // Function to find the sum of previous // numbers that are greater than the // current number for the given array static void sumGreater( int []ar, int N) { // Loop to iterate over all // the elements of the array for ( int i = 0; i < N; i++) { // Store the answer for // the current element int cur_sum = 0; // Iterate from (current index - 1) // to 0 and check if ar[j] is greater // than the current element and add // it to the cur_sum if so for ( int j = i - 1; j >= 0; j--) { if (ar[j] > ar[i]) cur_sum += ar[j]; } // Print the answer for // current element Console.Write(cur_sum + " " ); } } // Driver Code public static void Main(String[] args) { // Given array []arr int []ar = { 7, 3, 6, 2, 1 }; // Size of the array int N = ar.Length; // Function call sumGreater(ar, N); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript program for the above approach // Max Element of the Array var maxn = 1000000; // Function to find the sum of previous // numbers that are greater than the // current number for the given array function sumGreater(ar, N) { // Loop to iterate over all // the elements of the array for (i = 0; i < N; i++) { // Store the answer for // the current element var cur_sum = 0; // Iterate from (current index - 1) // to 0 and check if ar[j] is greater // than the current element and add // it to the cur_sum if so for (j = i - 1; j >= 0; j--) { if (ar[j] > ar[i]) cur_sum += ar[j]; } // Print the answer for // current element document.write(cur_sum + " " ); } } // Driver Code // Given array arr var ar = [ 7, 3, 6, 2, 1 ]; // Size of the array var N = ar.length; // Function call sumGreater(ar, N); // This code is contributed by umadevi9616 </script> |
0 7 7 16 18
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach the idea is to use Fenwick Tree. Below are the steps:
- Traverse the given array and find the sum(say total_sum) of all the elements stored in the Fenwick Tree.
- Now Consider each element(say arr[i]) as the index of the Fenwick Tree.
- Now find the sum of all the elements(say curr_sum) which is smaller than the current element using values stored in Tree.
- The value of total_sum – curr_sum will give the sum of all elements which are strictly greater than the elements on the left side of the current element.
- Update the current element in the Fenwick Tree.
- Repeat the above steps for all the elements in the array.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Max Element of the Array const int maxn = 1000000; // Initializing Fenwick Tree int Bit[maxn + 5]; // Function to calculate the sum of // previous numbers that are greater // than the current number in the array void sum( int ar[], int N) { // Iterate from 1 to N for ( int i = 0; i < N; i++) { int index; int total_sum = 0; index = 100000; // If some greater values has // occurred before current element // then it will be already stored // in Fenwick Tree while (index) { // Calculating sum of // all the elements total_sum += Bit[index]; index -= index & -index; } int cur_sum = 0; // Sum only smaller or equal // elements than current element index = ar[i]; while (index) { // If some smaller values has // occurred before it will be // already stored in Tree cur_sum += Bit[index]; index -= (index & -index); } int ans = total_sum - cur_sum; cout << ans << " " ; // Update the fenwick tree index = ar[i]; while (index <= 100000) { // Updating The Fenwick Tree // for future values Bit[index] += ar[i]; index += (index & -index); } } } // Driver Code int main() { // Given array arr[] int ar[] = { 7, 3, 6, 2, 1 }; int N = sizeof ar / sizeof ar[0]; // Function call sum(ar, N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Max Element of the Array static int maxn = 1000000 ; // Initializing Fenwick Tree static int []Bit = new int [maxn + 5 ]; // Function to calculate the sum of // previous numbers that are greater // than the current number in the array static void sum( int ar[], int N) { // Iterate from 1 to N for ( int i = 0 ; i < N; i++) { int index; int total_sum = 0 ; index = 100000 ; // If some greater values has // occurred before current element // then it will be already stored // in Fenwick Tree while (index > 0 ) { // Calculating sum of // all the elements total_sum += Bit[index]; index -= index & -index; } int cur_sum = 0 ; // Sum only smaller or equal // elements than current element index = ar[i]; while (index > 0 ) { // If some smaller values has // occurred before it will be // already stored in Tree cur_sum += Bit[index]; index -= (index & -index); } int ans = total_sum - cur_sum; System.out.print(ans + " " ); // Update the fenwick tree index = ar[i]; while (index <= 100000 ) { // Updating The Fenwick Tree // for future values Bit[index] += ar[i]; index += (index & -index); } } } // Driver Code public static void main(String[] args) { // Given array arr[] int ar[] = { 7 , 3 , 6 , 2 , 1 }; int N = ar.length; // Function call sum(ar, N); } } // This code is contributed by Rohit_ranjan |
Python3
# Python3 program for the above approach # Max Element of the Array maxn = 1000000 ; # Initializing Fenwick Tree Bit = [ 0 ] * (maxn + 5 ); # Function to calculate the sum of # previous numbers that are greater # than the current number in the array def sum (ar, N): # Iterate from 1 to N for i in range (N): total_sum = 0 ; index = 100000 ; # If some greater values has # occurred before current element # then it will be already stored # in Fenwick Tree while (index > 0 ): # Calculating sum of # all the elements total_sum + = Bit[index]; index - = index & - index; cur_sum = 0 ; # Sum only smaller or equal # elements than current element index = ar[i]; while (index > 0 ): # If some smaller values has # occurred before it will be # already stored in Tree cur_sum + = Bit[index]; index - = (index & - index); ans = total_sum - cur_sum; print (ans, end = " " ); # Update the fenwick tree index = ar[i]; while (index < = 100000 ): # Updating The Fenwick Tree # for future values Bit[index] + = ar[i]; index + = (index & - index); # Driver Code if __name__ = = '__main__' : # Given array arr arr = [ 7 , 3 , 6 , 2 , 1 ]; N = len (arr); # Function call sum (arr, N); # This code is contributed by sapnasingh4991 |
C#
// C# program for the above approach using System; class GFG{ // Max Element of the Array static int maxn = 1000000; // Initializing Fenwick Tree static int []Bit = new int [maxn + 5]; // Function to calculate the sum of // previous numbers that are greater // than the current number in the array static void sum( int []ar, int N) { // Iterate from 1 to N for ( int i = 0; i < N; i++) { int index; int total_sum = 0; index = 100000; // If some greater values has // occurred before current element // then it will be already stored // in Fenwick Tree while (index > 0) { // Calculating sum of // all the elements total_sum += Bit[index]; index -= index & -index; } int cur_sum = 0; // Sum only smaller or equal // elements than current element index = ar[i]; while (index > 0) { // If some smaller values has // occurred before it will be // already stored in Tree cur_sum += Bit[index]; index -= (index & -index); } int ans = total_sum - cur_sum; Console.Write(ans + " " ); // Update the fenwick tree index = ar[i]; while (index <= 100000) { // Updating The Fenwick Tree // for future values Bit[index] += ar[i]; index += (index & -index); } } } // Driver Code public static void Main(String[] args) { // Given array []arr int []ar = { 7, 3, 6, 2, 1 }; int N = ar.Length; // Function call sum(ar, N); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // javascript program for the above approach // Max Element of the Array var maxn = 1000000; // Initializing Fenwick Tree var Bit = Array(maxn + 5).fill(0); // Function to calculate the sum of // previous numbers that are greater // than the current number in the array function sum(ar , N) { // Iterate from 1 to N for (i = 0; i < N; i++) { var index; var total_sum = 0; index = 100000; // If some greater values has // occurred before current element // then it will be already stored // in Fenwick Tree while (index > 0) { // Calculating sum of // all the elements total_sum += Bit[index]; index -= index & -index; } var cur_sum = 0; // Sum only smaller or equal // elements than current element index = ar[i]; while (index > 0) { // If some smaller values has // occurred before it will be // already stored in Tree cur_sum += Bit[index]; index -= (index & -index); } var ans = total_sum - cur_sum; document.write(ans + " " ); // Update the fenwick tree index = ar[i]; while (index <= 100000) { // Updating The Fenwick Tree // for future values Bit[index] += ar[i]; index += (index & -index); } } } // Driver Code // Given array arr var ar = [ 7, 3, 6, 2, 1 ]; var N = ar.length; // Function call sum(ar, N); // This code contributed by aashish1995 </script> |
0 7 7 16 18
Time Complexity: O(N * log(max_element))
Auxiliary Space: O(max_element)
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